Search Results

Search found 1181 results on 48 pages for 'letters'.

Page 7/48 | < Previous Page | 3 4 5 6 7 8 9 10 11 12 13 14  | Next Page >

  • Reading a user input (character or string of letters) into ggplot command inside a switch statement or a nested ifelse (with functions in it)

    - by statisticalbeginner
    I have code like AA <- as.integer(readline("Select any number")) switch(AA, 1={ num <-as.integer(readline("Select any one of the options \n")) print('You have selected option 1') #reading user data var <- readline("enter the variable name \n") #aggregating the data based on required condition gg1 <- aggregate(cbind(get(var))~Mi+hours,a, FUN=mean) #Ploting ggplot(gg1, aes(x = hours, y = get(var), group = Mi, fill = Mi, color = Mi)) + geom_point() + geom_smooth(stat="smooth", alpha = I(0.01)) }, 2={ print('bar') }, { print('default') } ) The dataset is [dataset][1] I have loaded the dataset into object list a <- read.table(file.choose(), header=FALSE,col.names= c("Ei","Mi","hours","Nphy","Cphy","CHLphy","Nhet","Chet","Ndet","Cdet","DON","DOC","DIN","DIC","AT","dCCHO","TEPC","Ncocco","Ccocco","CHLcocco","PICcocco","par","Temp","Sal","co2atm","u10","dicfl","co2ppm","co2mol","pH")) I am getting error like source ("switch_statement_check.R") Select any one of the options 1 [1] "You have selected option 1" enter the variable name Nphy Error in eval(expr, envir, enclos) : (list) object cannot be coerced to type 'double' > gg1 is getting data that is fine. I dont know what to do to make the variable entered by user to work in that ggplot command. Please suggest any solution for this. The dput output structure(list(Ei = c(1L, 1L, 1L, 1L, 1L, 1L), Mi = c(1L, 1L, 1L, 1L, 1L, 1L), hours = 1:6, Nphy = c(0.1023488, 0.104524, 0.1064772, 0.1081702, 0.1095905, 0.110759), Cphy = c(0.6534707, 0.6448216, 0.6369597, 0.6299084, 0.6239005, 0.6191941), CHLphy = c(0.1053458, 0.110325, 0.1148174, 0.1187672, 0.122146, 0.1249877), Nhet = c(0.04994161, 0.04988347, 0.04982555, 0.04976784, 0.04971029, 0.04965285), Chet = c(0.3308593, 0.3304699, 0.3300819, 0.3296952, 0.3293089, 0.3289243), Ndet = c(0.04991916, 0.04984045, 0.04976363, 0.0496884, 0.04961446, 0.04954156), Cdet = c(0.3307085, 0.3301691, 0.3296314, 0.3290949, 0.3285598, 0.3280252), DON = c(0.05042275, 0.05085697, 0.05130091, 0.05175249, 0.05220978, 0.05267118 ), DOC = c(49.76304, 49.52745, 49.29323, 49.06034, 48.82878, 48.59851), DIN = c(14.9933, 14.98729, 14.98221, 14.9781, 14.97485, 14.97225), DIC = c(2050.132, 2050.264, 2050.396, 2050.524, 2050.641, 2050.758), AT = c(2150.007, 2150.007, 2150.007, 2150.007, 2150.007, 2150.007), dCCHO = c(0.964222, 0.930869, 0.8997098, 0.870544, 0.843196, 0.8175117), TEPC = c(0.1339044, 0.1652179, 0.1941872, 0.2210289, 0.2459341, 0.2690721), Ncocco = c(0.1040715, 0.1076058, 0.1104229, 0.1125141, 0.1140222, 0.1151228), Ccocco = c(0.6500288, 0.6386706, 0.6291149, 0.6213265, 0.6152447, 0.6108502), CHLcocco = c(0.1087667, 0.1164099, 0.1225822, 0.1273103, 0.1308843, 0.1336465), PICcocco = c(0.1000664, 0.1001396, 0.1007908, 0.101836, 0.1034179, 0.1055634), par = c(0, 0, 0.8695131, 1.551317, 2.777707, 4.814341), Temp = c(9.9, 9.9, 9.9, 9.9, 9.9, 9.9), Sal = c(31.31, 31.31, 31.31, 31.31, 31.31, 31.31), co2atm = c(370, 370, 370, 370, 370, 370), u10 = c(0.01, 0.01, 0.01, 0.01, 0.01, 0.01), dicfl = c(-2.963256, -2.971632, -2.980446, -2.989259, -2.997877, -3.005702), co2ppm = c(565.1855, 565.7373, 566.3179, 566.8983, 567.466, 567.9814), co2mol = c(0.02562326, 0.02564828, 0.0256746, 0.02570091, 0.02572665, 0.02575002 ), pH = c(7.879427, 7.879042, 7.878636, 7.878231, 7.877835, 7.877475)), .Names = c("Ei", "Mi", "hours", "Nphy", "Cphy", "CHLphy", "Nhet", "Chet", "Ndet", "Cdet", "DON", "DOC", "DIN", "DIC", "AT", "dCCHO", "TEPC", "Ncocco", "Ccocco", "CHLcocco", "PICcocco", "par", "Temp", "Sal", "co2atm", "u10", "dicfl", "co2ppm", "co2mol", "pH"), row.names = c(NA, 6L), class = "data.frame") As per the below suggestions I have tried a lot but it is not working. Summarizing I will say: var <- readline("enter a variable name") I cant use get(var) inside any command but not inside ggplot, it wont work. gg1$var it also doesnt work, even after changing the column names. Does it have a solution or should I just choose to import from an excel sheet, thats better? Tried with if else and functions fun1 <- function() { print('You have selected option 1') my <- as.character((readline("enter the variable name \n"))) gg1 <- aggregate(cbind(get(my))~Mi+hours,a, FUN=mean) names(gg1)[3] <- my #print(names(gg1)) ggplot (gg1,aes_string(x="hours",y=(my),group="Mi",color="Mi")) + geom_point() } my <- as.integer(readline("enter a number")) ifelse(my == 1,fun1(),"") ifelse(my == 2,print ("its 2"),"") ifelse(my == 3,print ("its 3"),"") ifelse(my != (1 || 2|| 3) ,print("wrong number"),"") Not working either...:(

    Read the article

  • Fastest way to get array of NSRange objects for all uppercase letters in an NSString?

    - by Bama91
    I need NSRange objects for the position of each uppercase letter in a given NSString for input into a method for a custom attributed string class.  There are of course quite a few ways to accomplish this such as rangeOfString:options: with NSRegularExpressionSearch or using RegexKitLite to get each match separately while walking the string.  What would be the fastest performing approach to accomplish this task?

    Read the article

  • How to prevent users from inputting letters or numbers ?

    - by ZaZu
    Hello, I have a simple problem; Here is the code : #include<stdio.h> main(){ int input; printf("Choose a numeric value"); scanf("%d",&input); } I want the user to only enter numbers ... So it has to be something like this : #include<stdio.h> main(){ int input; printf("Choose a numeric value"); do{ scanf("%d",&input); }while(input!= 'something'); } My problem is that I dont know what to replace in 'something' ... How can I prevent users from inputting alphabetic characters ? Thanks for your help ! }

    Read the article

  • Printing Multiple html page through ASP.NET

    - by sameer
    Hi All, I have multiple letters in html format in a folder and im list those letters in the gridview on ASP.NET page if user select couple of letters and click print i should me those letters Now Question is how to print multiple html files on the event of button click.

    Read the article

  • Javascript regular expressions: how to match ONLY the given characters?

    - by Dfowj
    I'm trying to use a regex like /[computer]{3,8}/ to get any words containing only the letters in computer ranging from 3 to 8 letters long. That regex instead captures all words containing ANY of the letters in [computer]. I've been looking at regular expression examples, but i can't quite figure it out... How do i modifiy this regular expression to capture words containing ONLY the letters in computer (with a length of 3 to 8)?

    Read the article

  • C++ - How to efficiently find out if any string in a vector can be assembled from a set of letters

    - by Francisco P.
    Hello, everyone! I am implementing a text-based version of Scrabble for a college project. I have a vector containing around 400K strings (my dictionary), and, at some point in every turn, I'm going to have to check if there's still a word in the dictionary which can be formed with the pieces in the player's hand. I'm checking if the player has any move left... If not, it's game over for the player in question... My only solution to this is iterating through the string, one by one, and using a sub-routine I have to check if the string in question can be formed from the player's pieces. I'll implement a quickfail checking if the user has any vowels, but it'll still be woefully inefficient. Any suggestions? Thanks for your time!

    Read the article

  • How to index small words (3 letters) with SQL Full-text search?

    - by Sly
    I have an Incident table with one row that has the value 'out of office' in the Description column. However the following query does not return that row. SELECT * FROM Incident WHERE CONTAINS( (Incident.Description), '"out*"' ) The word 'out' is not in the noise file (I cleared the noise file completely). Is it because SQL Full-text search does not index small words? Is there a setting for that? Note: I'm on SQL 2005.

    Read the article

  • hide taskbar labels completely. no letters no words in the open windows. like windows 7, icons only

    - by Dom
    i posted this question in an answer box, so once person had to tell me to make it a separate question. http://superuser.com/questions/30007/hide-taskbar-labels-without-combining My Question: it shows 1 letter. i just want plain icons. like that of windows 7. no words/letter. only icons. how do u get rid of the letter? btw im using xp, i just want no words no letters on the taskbar. is it possible? answers appreciated(except bashful...)!

    Read the article

  • Python: How would i write this 'if' statement for a word of arbitrary length?

    - by ElCarlos
    This is what I currently have: wordlist = [fox, aced, definite, ace] for word in wordlist: a = len(word) if (ord(word[a-(a-1)] - ord(word[(a-a)])) == ord(word[a-(a-2)])-ord(word[a-(a-1)]: print "success", word else: print "fail", word What I'm trying to do is calculate the ASCII values between each of the letters in the word. And check to see if the ord of the letters are increasing by the same value. so for fox, it would check if the difference between the ord of 2nd and 1st letters are equal to the ord difference of the 3rd and 2nd letters. However, with my current 'if' statement, only the first 3 letters of a word are compared. How can I rewrite this statement to cover every letter in a word of length greater than 3? Sorry if I can't present this clearly, thanks for your time.

    Read the article

  • All words in a trie data-structure

    - by John Smith
    I'm trying to put all words in a trie in a string, a word is detonated by the eow field being true for a certain character in the trie data structure, hence a trie can could have letters than lead up to no word, for ex "abc" is in the trie but "c"'s eow field is false so "abc" is not a word Here is my Data structure struct Trie { bool eow; //when a Trie field isWord = true, hence there is a word char letter; Trie *letters[27]; }; and here is my attemped print-all function, basically trying to return all words in one string seperated by spaces for words string printAll( string word, Trie& data) { if (data.eow == 1) return word + " "; for (int i = 0; i < 26; i++) { if (data.letters[i] != NULL) printAll( word + data.letters[i]->letter, *(data.letters[i])); } return ""; } Its not outputting what i want, any suggestions?

    Read the article

  • Python: How can I read in the characters from a line in a file and convert them to floats and strs, depending on if they are numbers or letters?

    - by user1467577
    I have a file that looks like: 1 1 C C 1.9873 2.347 3.88776 1 2 C Si 4.887 9.009 1.21 I would like to read in the contents of the file, line-by-line. When I only had numbers on the lines I used: for line in readlines(file): data = map(float, line.split) But this only works when all the elements of line.split are numbers. How can I make it store the letters as strings and the numbers as floats?

    Read the article

  • SQL SERVER – Puzzle #1 – Querying Pattern Ranges and Wild Cards

    - by Pinal Dave
    Note: Read at the end of the blog post how you can get five Joes 2 Pros Book #1 and a surprise gift. I have been blogging for almost 7 years and every other day I receive questions about Querying Pattern Ranges. The most common way to solve the problem is to use Wild Cards. However, not everyone knows how to use wild card properly. SQL Queries 2012 Joes 2 Pros Volume 1 – The SQL Queries 2012 Hands-On Tutorial for Beginners Book On Amazon | Book On Flipkart Learn SQL Server get all the five parts combo kit Kit on Amazon | Kit on Flipkart Many people know wildcards are great for finding patterns in character data. There are also some special sequences with wildcards that can give you even more power. This series from SQL Queries 2012 Joes 2 Pros® Volume 1 will show you some of these cool tricks. All supporting files are available with a free download from the www.Joes2Pros.com web site. This example is from the SQL 2012 series Volume 1 in the file SQLQueries2012Vol1Chapter2.2Setup.sql. If you need help setting up then look in the “Free Videos” section on Joes2Pros under “Getting Started” called “How to install your labs” Querying Pattern Ranges The % wildcard character represents any number of characters of any length. Let’s find all first names that end in the letter ‘A’. By using the percentage ‘%’ sign with the letter ‘A’, we achieve this goal using the code sample below: SELECT * FROM Employee WHERE FirstName LIKE '%A' To find all FirstName values beginning with the letters ‘A’ or ‘B’ we can use two predicates in our WHERE clause, by separating them with the OR statement. Finding names beginning with an ‘A’ or ‘B’ is easy and this works fine until we want a larger range of letters as in the example below for ‘A’ thru ‘K’: SELECT * FROM Employee WHERE FirstName LIKE 'A%' OR FirstName LIKE 'B%' OR FirstName LIKE 'C%' OR FirstName LIKE 'D%' OR FirstName LIKE 'E%' OR FirstName LIKE 'F%' OR FirstName LIKE 'G%' OR FirstName LIKE 'H%' OR FirstName LIKE 'I%' OR FirstName LIKE 'J%' OR FirstName LIKE 'K%' The previous query does find FirstName values beginning with the letters ‘A’ thru ‘K’. However, when a query requires a large range of letters, the LIKE operator has an even better option. Since the first letter of the FirstName field can be ‘A’, ‘B’, ‘C’, ‘D’, ‘E’, ‘F’, ‘G’, ‘H’, ‘I’, ‘J’ or ‘K’, simply list all these choices inside a set of square brackets followed by the ‘%’ wildcard, as in the example below: SELECT * FROM Employee WHERE FirstName LIKE '[ABCDEFGHIJK]%' A more elegant example of this technique recognizes that all these letters are in a continuous range, so we really only need to list the first and last letter of the range inside the square brackets, followed by the ‘%’ wildcard allowing for any number of characters after the first letter in the range. Note: A predicate that uses a range will not work with the ‘=’ operator (equals sign). It will neither raise an error, nor produce a result set. --Bad query (will not error or return any records) SELECT * FROM Employee WHERE FirstName = '[A-K]%' Question: You want to find all first names that start with the letters A-M in your Customer table and end with the letter Z. Which SQL code would you use? a. SELECT * FROM Customer WHERE FirstName LIKE 'm%z' b. SELECT * FROM Customer WHERE FirstName LIKE 'a-m%z' c. SELECT * FROM Customer WHERE FirstName LIKE 'a-m%z' d. SELECT * FROM Customer WHERE FirstName LIKE '[a-m]%z' e. SELECT * FROM Customer WHERE FirstName LIKE '[a-m]z%' f. SELECT * FROM Customer WHERE FirstName LIKE '[a-m]%z' g. SELECT * FROM Customer WHERE FirstName LIKE '[a-m]z%' Contest Leave a valid answer before June 18, 2013 in the comment section. 5 winners will be selected from all the valid answers and will receive Joes 2 Pros Book #1. 1 Lucky person will get a surprise gift from Joes 2 Pros. The contest is open for all the countries where Amazon ships the book (USA, UK, Canada, India and many others). Special Note: Read all the options before you provide valid answer as there is a small trick hidden in answers. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: Joes 2 Pros, PostADay, SQL, SQL Authority, SQL Puzzle, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

    Read the article

  • Acting as though the laptop Fn key is pressed

    - by Matt
    I have a Fujitsu laptop and all was working fine until I ended teamviewer. On the keyboard it has a numberpad on some of the letters which you press the Fn button to use. The system is acting like I have the Fn key pressed and I get numbers shown when it should be letters. If I press the Fn key the numbers go to letters. The Fn key is not stuck down. Is there a way to reset keyboard settings or a known fix. The operating system is Ubuntu 11.04 Natty.

    Read the article

  • How can I estimate the entropy of a password?

    - by Wug
    Having read various resources about password strength I'm trying to create an algorithm that will provide a rough estimation of how much entropy a password has. I'm trying to create an algorithm that's as comprehensive as possible. At this point I only have pseudocode, but the algorithm covers the following: password length repeated characters patterns (logical) different character spaces (LC, UC, Numeric, Special, Extended) dictionary attacks It does NOT cover the following, and SHOULD cover it WELL (though not perfectly): ordering (passwords can be strictly ordered by output of this algorithm) patterns (spatial) Can anyone provide some insight on what this algorithm might be weak to? Specifically, can anyone think of situations where feeding a password to the algorithm would OVERESTIMATE its strength? Underestimations are less of an issue. The algorithm: // the password to test password = ? length = length(password) // unique character counts from password (duplicates discarded) uqlca = number of unique lowercase alphabetic characters in password uquca = number of uppercase alphabetic characters uqd = number of unique digits uqsp = number of unique special characters (anything with a key on the keyboard) uqxc = number of unique special special characters (alt codes, extended-ascii stuff) // algorithm parameters, total sizes of alphabet spaces Nlca = total possible number of lowercase letters (26) Nuca = total uppercase letters (26) Nd = total digits (10) Nsp = total special characters (32 or something) Nxc = total extended ascii characters that dont fit into other categorys (idk, 50?) // algorithm parameters, pw strength growth rates as percentages (per character) flca = entropy growth factor for lowercase letters (.25 is probably a good value) fuca = EGF for uppercase letters (.4 is probably good) fd = EGF for digits (.4 is probably good) fsp = EGF for special chars (.5 is probably good) fxc = EGF for extended ascii chars (.75 is probably good) // repetition factors. few unique letters == low factor, many unique == high rflca = (1 - (1 - flca) ^ uqlca) rfuca = (1 - (1 - fuca) ^ uquca) rfd = (1 - (1 - fd ) ^ uqd ) rfsp = (1 - (1 - fsp ) ^ uqsp ) rfxc = (1 - (1 - fxc ) ^ uqxc ) // digit strengths strength = ( rflca * Nlca + rfuca * Nuca + rfd * Nd + rfsp * Nsp + rfxc * Nxc ) ^ length entropybits = log_base_2(strength) A few inputs and their desired and actual entropy_bits outputs: INPUT DESIRED ACTUAL aaa very pathetic 8.1 aaaaaaaaa pathetic 24.7 abcdefghi weak 31.2 H0ley$Mol3y_ strong 72.2 s^fU¬5ü;y34G< wtf 88.9 [a^36]* pathetic 97.2 [a^20]A[a^15]* strong 146.8 xkcd1** medium 79.3 xkcd2** wtf 160.5 * these 2 passwords use shortened notation, where [a^N] expands to N a's. ** xkcd1 = "Tr0ub4dor&3", xkcd2 = "correct horse battery staple" The algorithm does realize (correctly) that increasing the alphabet size (even by one digit) vastly strengthens long passwords, as shown by the difference in entropy_bits for the 6th and 7th passwords, which both consist of 36 a's, but the second's 21st a is capitalized. However, they do not account for the fact that having a password of 36 a's is not a good idea, it's easily broken with a weak password cracker (and anyone who watches you type it will see it) and the algorithm doesn't reflect that. It does, however, reflect the fact that xkcd1 is a weak password compared to xkcd2, despite having greater complexity density (is this even a thing?). How can I improve this algorithm? Addendum 1 Dictionary attacks and pattern based attacks seem to be the big thing, so I'll take a stab at addressing those. I could perform a comprehensive search through the password for words from a word list and replace words with tokens unique to the words they represent. Word-tokens would then be treated as characters and have their own weight system, and would add their own weights to the password. I'd need a few new algorithm parameters (I'll call them lw, Nw ~= 2^11, fw ~= .5, and rfw) and I'd factor the weight into the password as I would any of the other weights. This word search could be specially modified to match both lowercase and uppercase letters as well as common character substitutions, like that of E with 3. If I didn't add extra weight to such matched words, the algorithm would underestimate their strength by a bit or two per word, which is OK. Otherwise, a general rule would be, for each non-perfect character match, give the word a bonus bit. I could then perform simple pattern checks, such as searches for runs of repeated characters and derivative tests (take the difference between each character), which would identify patterns such as 'aaaaa' and '12345', and replace each detected pattern with a pattern token, unique to the pattern and length. The algorithmic parameters (specifically, entropy per pattern) could be generated on the fly based on the pattern. At this point, I'd take the length of the password. Each word token and pattern token would count as one character; each token would replace the characters they symbolically represented. I made up some sort of pattern notation, but it includes the pattern length l, the pattern order o, and the base element b. This information could be used to compute some arbitrary weight for each pattern. I'd do something better in actual code. Modified Example: Password: 1234kitty$$$$$herpderp Tokenized: 1 2 3 4 k i t t y $ $ $ $ $ h e r p d e r p Words Filtered: 1 2 3 4 @W5783 $ $ $ $ $ @W9001 @W9002 Patterns Filtered: @P[l=4,o=1,b='1'] @W5783 @P[l=5,o=0,b='$'] @W9001 @W9002 Breakdown: 3 small, unique words and 2 patterns Entropy: about 45 bits, as per modified algorithm Password: correcthorsebatterystaple Tokenized: c o r r e c t h o r s e b a t t e r y s t a p l e Words Filtered: @W6783 @W7923 @W1535 @W2285 Breakdown: 4 small, unique words and no patterns Entropy: 43 bits, as per modified algorithm The exact semantics of how entropy is calculated from patterns is up for discussion. I was thinking something like: entropy(b) * l * (o + 1) // o will be either zero or one The modified algorithm would find flaws with and reduce the strength of each password in the original table, with the exception of s^fU¬5ü;y34G<, which contains no words or patterns.

    Read the article

  • Regex to match card code input

    - by kate
    How can I write a regex to match strings following these rules? 1 letter followed by 4 letters or numbers, then 5 letters or numbers, then 3 letters or numbers followed by a number and one of the following signs: ! & @ ? I need to allow input as a 15-character string or as 3 groups of 5 chars separated by one space. I'm implementing this in JavaScript.

    Read the article

  • Regex Question ...

    - by kate
    Hi, Could someone help me with the following RegEx query: based on the following rules: 1) 1 letter followed by 4 letters or numbers, then 2) 5 letters or numbers, then 3) 3 letters or numbers followed by a number and one of the following signs: ! & @ ? You will have to allow customers to input the fidelity card code as a 15-character string, or as 3 groups of 5 chars, separated by one space.

    Read the article

  • typeahead.js remote with subset matching

    - by rebelde
    Instead of returning to the server after each additional letter is typed, I want it to only go to the server once, get all matching words, and filter the downloaded data after that. We are having trouble making this work. We are successfully using "remote" to wait until two letters are typed, but we can't get it to stop going to the server as additional letters are typed. Steps: 1. After two letters are typed, retrieve all matching words that start with those two letters. 2. When a third and additional letters are typed, don't go to the server again, just filter from the previous list that was sent. An example: "mo" is typed in. All 100 words that start with "mo" are returned. (Only 10 are shown.) "mor" - now with a third letter, we don't go back to the server. We just find the 20 words that match from within the previous set of words. Can anybody make this work? In real life (using YUI2), we do this and then go back to the server if somebody types in a space after the word. At that point, we know to retrieve additional words. Thanks!

    Read the article

  • IE6 is duplicating characters, can't figure out why. Suggestions?

    - by Paul
    Problem is located on http://www.preownedweddingdresses.com/ We have a dresses slider at the bottom, select tabs different dresses shown. Works fine everywhere else, but for some reason, in IE6, the letters "ls" (from the tab "Best Deals") are duplicating inside the content and causing rendering issues. I've yet to find anything that can fix this, or anything that can be blamed for causing this either. I've changed the letters at the end of Best Deals, and the duplicated letters change as well. Open to any suggestions.

    Read the article

  • concatenate arbitrary long list of matches in SQL subquery

    - by lordvlad
    imagine 2 tables (rather stupid example, but for the sake of simplicity, here you go) words word_id letters letter word_id how can i select all words while selecting all letters that belong to a word and concatenating them to said word? it is important that the letters are returned in the order they appear in the table, as the letter may be mixed into other words, but the order is correct. |word_id| |word_id|letter| +-------+ +-------+------+ | 1| | 1| H| | 2| | 2| B| | 2| Y| | 1| I| | 2| E| should return |word_id|word| +-------+----+ | 1| HI| | 2| BYE| any way to accomplish this in pure SQL?

    Read the article

  • Removing Item From List - during iteration - what's wrong with this idiom ?

    - by monojohnny
    As an experiment, I did this: letters=['a','b','c','d','e','f','g','h','i','j','k','l'] for i in letters: letters.remove(i) print letters The last print shows that not all items were removed ? (every other was). IDLE 2.6.2 >>> ================================ RESTART ================================ >>> ['b', 'd', 'f', 'h', 'j', 'l'] >>> What's the explanation for this ? How it could this be re-written to remove every item ?

    Read the article

  • C# A simple Hangman game

    - by Radostin Angelov
    I'm trying to create a simple Hangman game and i have gotten so far, to make it read all words from a text file, but I don't know how to make the code work for every single word. I have another project, working with 3/4 words but with repeating nested if statements. I want to make it as shorter as possible. This is the code i have so far : using System; using System.Linq; class Program { static void Main() { string[] words = System.IO.File.ReadAllLines(@"C:\Users\ADMIN\Desktop\Letters\Letters.txt"); int LengthOfArray = words.Length; Random rnd = new Random(); int random = rnd.Next(1, 3); char[] letters = words[random].ToCharArray(); bool WordIsHidden = true; char hiddenChar = '_'; char GuessedLetter = hiddenChar; var retry = true; while (retry = true) { Console.WriteLine(letters); letters = GuessedLetter.ToString().ToCharArray(); for (int i = 1; i <= LengthOfArray; i++) { Console.Write("{0} ", GuessedLetter); } Console.WriteLine("Enter a letter!"); char letter = char.Parse(Console.ReadLine()); if (words[random].Contains<char>(letter)) { WordIsHidden = false; GuessedLetter = letter; Console.Write(letters); } else { if (WordIsHidden == true) { Console.Write("You guessed wrong!"); } } } } } Also I'm trying to make the game show each letter, the user has guessed on it's corresponding position, but now the letter is one line higher than the rest of the word and it's not in it's right position. Edited: Here is the result : cat ___Enter a letter! a __ aaaEnter a letter! t aa tttEnter a letter! IF anyone have a clue for where does this come from and how can I fix it, any help will be greatly appreciated.

    Read the article

< Previous Page | 3 4 5 6 7 8 9 10 11 12 13 14  | Next Page >